Problem: A rectangular picture frame is made from one-inch-wide pieces of wood.  The area of just the frame is $18$ square inches, and one of the outer edges of the frame is $5$ inches long.  What is the sum of the lengths of the four interior edges of the frame?

[asy]

size(5cm,5cm);

draw((0,0)--(5,0)--(5,7)--(0,7)--(0,0));

draw((1,1)--(4,1)--(4,6)--(1,6)--(1,1));

fill(((1,1)--(4,1)--(4,6)--(1,6)--cycle),darkblue);

draw (shift(0, 0.5)*((0,7)--(5,7)), Bars);

label("$5''$",(2.5,7.5),N);

draw (shift(0, -0.5)*((4,0)--(5,0)), Bars);

label("$1''$",(4.5,-0.5),S);

draw (shift(0.5,0)*((5,0)--(5,1)), Bars);

label("$1''$",(5.5,0.5),E);

[/asy]
Solution: The lengths of the top/bottom interior edges are $5-2=3$ inches (since there is 1 inch of frame to either side of the interior rectangular hole). Let the lengths of the left/right interior edges be $x$ inches. Then, the lengths of the left/right exterior edges are $x+2$ inches. The area of the frame is equal to the area of the frame rectangle minus the area of the hole. This is equal to $5\cdot(x+2)-3x=2x+10$. Since we are given that this area is 18 square inches, we have the equation $2x+10=18\Rightarrow x=4$. So the dimensions of the interior hole are $3\times4$. Thus, the sum of the four interior edges is $3+4+3+4=\boxed{14}$ inches.